Problem: $81$ people attended a baseball game. Everyone there was a fan of either the home team or the away team. The number of home team fans was $99$ less than $4$ times the number of away team fans. How many home team and away team fans attended the game?
Solution: Let $x$ equal the number of home team fans and $y$ equal the number of away team fans. The system of equations is then: ${x+y = 81}$ ${x = 4y-99}$ Solve for $x$ and $y$ using substitution. Since $x$ has already been solved for, substitute ${4y-99}$ for $x$ in the first equation. ${(4y-99)}{+ y = 81}$ Simplify and solve for $y$ $ 4y-99 + y = 81 $ $ 5y-99 = 81 $ $ 5y = 180 $ $ y = \dfrac{180}{5} $ ${y = 36}$ Now that you know ${y = 36}$ , plug it back into ${x = 4y-99}$ to find $x$ ${x = 4}{(36)}{ - 99}$ $x = 144 - 99$ ${x = 45}$ You can also plug ${y = 36}$ into ${x+y = 81}$ and get the same answer for $x$ ${x + }{(36)}{= 81}$ ${x = 45}$ There were $45$ home team fans and $36$ away team fans.